Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *.
Example 1
Input: "2-1-1".
((2-1)-1) = 0 (2-(1-1)) = 2
Output: [0, 2]
Example 2
Input: "2*3-4*5"
(2*(3-(4*5))) = -34 ((2*3)-(4*5)) = -14 ((2*(3-4))*5) = -10 (2*((3-4)*5)) = -10 (((2*3)-4)*5) = 10
Output: [-34, -14, -10, -10, 10]
实现；
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
for (int i = 0; i < input.size(); i++) {
char ch = input[i];
if (ch == ‘+‘ || ch == ‘-‘ || ch == ‘*‘) {
vector<int> lv = diffWaysToCompute(input.substr(0, i));
vector<int> rv = diffWaysToCompute(input.substr(i+1));
for (auto x : lv) {
for (auto y : rv) {
if (ch == ‘+‘) {
result.push_back(x+y);
} else if (ch == ‘*‘) {
result.push_back(x*y);
} else if (ch == ‘-‘) {
result.push_back(x-y);
}
}
}
}
}
if (result.empty()) {
result.push_back(atoi(input.c_str()));
}
return result;
}
};
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